The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is
[Ag = 108, F = 96500 c mol-1]
- 108.0g
-
54.0g
- 27.0g
- 13.5gq
The correct answer is: B
Explanation
Ag\(^+\) + e\(^-\) β Ag
If 96500 C deposited 108g of Ag
β΄ 10 x 4830C will deposit Xg of Ag
X = \(\frac{108 x 48300}{96500}\)
X = 54.06g